I .T HEO CHirONG TRNH CHUAN
Do d6: V Jd NaOH •
19.22. Dap dn diing Ih B.
Cac phuang trinh phan ung:
2A1 + 3H2SO4 -> AlaCSOJ, + SHzt Zn + H2SO4 -> ZnS04 + H j t
Goi X, y la s6' mol A l va Zn c6 trong h6n hop.
Ta c6: 27x + 65y = 3,68
3 2 24
S6'molH,: - x + y = — = 0 , 1 ' 2 ^ 22,4
Giai he phuang trinh dugc: x = y = 0,04 mol.
T h e o ( l ) v a (2): m^i = 0,04.27 = 1,08 (gam); mz„ = 0,04.65 = 2,60 (gam). (1) (2) m H2SO4 -.0,04 + 0,04 2 .98-9,8 (gam)
^ Kh6'i l u g n g dung dich H2SO4 10% la 98 gam.
V i C O 0,2 gam bay ra, nen kh6'i lugfng d u n g d j c h t h u dugc sau
phan ling la: m = 98 + 1,08 + 2,60 - 0,2 = 101,48 (gam).
19.23. D a p an d i i n g la Ạ
Sau k h i phan ling xay ra hoan toan: .
- T r o n g d u n g dich c6 ba i o n k i m loai la: Mg^"", Zn^"" Cu^"" d u . - S6' mol dien t i c h d u a n g trudc va sau phan ling:
TrUffc phan Aug:
n^^^. = 2 mol s6' mol dien tfch + la 4 mol.
n . . = 1 mol -> so mol dien tfch + la 1 mol.
Ag
Sau phan i^ng:
"zn=* ~ ^ ~^ ^^^^ ^^^^ •'"
n^g> = 1,2 mol -> s6' mol dien tfch + la 2,4 mol. V a dien tfch + cua ion Cu^* dụ
Theo dinh luat bao toan dien tfch:
2x + 2,4 < 4 + 1
- > x < l, 3 .
V a y gia tri X = 1,2 Ik phii hop.
19.24. D a p an d i i n g la D .
19.25. D a p an d i i n g la B.
Phuong t r i n h phan ling:
K2Cr207 + 14HC1 ^ 2CrCl, + 3CI2 + 2KC1 + 7H2O
S6 phan tijf HCl d6ng vai tr6 chat khijf: k = 6 (6HC1 SCIj + 6H*) Tdng s6' phan tur HCl tham gia phan ling: t = 14.
k 6 3 t 14 7 |9.26. D a p an d i i n g la B. CuO, Fe203 (X) f H C l ^ CUCI2, 2FeCl3 (Y) (1)
Theo (1): K h i chuy^n X thanh Y da c6 hai CI thay thé m6t O (oxi) -> Kh6'i lucmg tang 71 - 16 = 55 (gam).
8 5 9 * 5 - 4 4
S6' mol O = -^'^^ ^ = 0,75 (mol) 55
CuO, Fe203 Cu, F e + C 0 2 t Theo (2): S6' mol COj bay ra = s6' mol O = 0,75 mol
CO2 + BăOH)2 -> B a C O , i + H2O Theo (3): nj^côi = "coj = Ó^S mol.
197.0,75
(2) (3)
m BaCO,i = 73,875 (gam).
|.27. Dap an diing la C.
Phuang trinh phan ling cua X vdi dung dich K O H du: Zn + 2H2O + K O H ^ K[Zn(OH),] + H j t 0,15 mol 0,15 mol nc„ = 0,25 - 0 , 1 5 = 0,1 (mol). nicu = 0,1.64 = 6,4 (gam), mx = 6,4 + 0,15.65 = 16,15 (gam). 6,4 (1) %mc„ = :.100% = 39,63%. 16,15 .28. Dap an diing la Ạ
Cac phuang trinh phan ling:
CuO + 2HCI ^ CUCI2 + H2O
CUCI2 + 2NH3 + 2H2O ^ Cu(0H)2>Í + 2NH4CI
Cu(OH)2 + 4NH3 [Cu(NH,)4](OH)2 (tan) (mau xanh thSm)
^9.29. Dap an diing la Ạ
A g N O , la mud'i nitrat ciia kim loai y6'u, khi bi nhiet phan tao thanh kim loai Ag, khf NO2 va khf O2.
19.30. Dap diing la C.